question_answer

If \[({{x}_{1}},\,\,{{y}_{1}})\] & \[({{x}_{2}},\,\,{{y}_{2}})\] are the ends of a diameter of a circle such that \[{{x}_{1}}\] & \[{{x}_{2}}\] are the roots of the equation \[a{{x}^{2}}+bx+c=0\] and \[\,{{y}_{1}}\] & \[{{y}_{2}}\] are the roots of the equation \[p{{y}^{2}}+qy+c=0.\] Then the co-ordinates of the centre of the circle is:

A) \[\left( \frac{b}{2a},\,\,\frac{q}{2p} \right)\]

B) \[\left( -\frac{b}{2a},\,\,-\frac{q}{2p} \right)\]

C) \[\left( \frac{b}{a},\,\,\frac{q}{p} \right)\]

D) none of these

Correct Answer: B

Solution :

\[{{x}_{1}}+{{x}_{2}}=-\,b/q\] and \[{{y}_{1}}+{{y}_{2}}=-\,q/p\]

Centre \[\left[ \frac{{{x}_{1}}+{{x}_{2}}}{2},\,\,\frac{{{y}_{1}}+{{y}_{2}}}{2} \right]\equiv \left[ -\frac{b}{2a},\,\,\frac{-\,q}{2p} \right]\]

[Hint: required equation is

\[{{x}^{2}}+{{y}^{2}}+\frac{b}{a}x+\frac{q}{p}y+\frac{c}{a}+\frac{c}{p}=0\]\[\Rightarrow \]B]